If cot θ = 7/8, evaluate : (i) (1 + sin θ)(1 - sin θ)/(1+cos θ)(1-cos θ) (ii) cot^2 θ

Dear Student

Let us assume
a△ABC in which∠B = 90° and∠C= θ
Given:
cotθ= BC/AB = 7/8
Let BC = 7k and AB = 8k,
where k is a positive real number
According to Pythagoras theorem in△ABC
we get

AC^2= AB^2+BC^2
AC^2= (8k)^2+(7k)^2
AC^2= 64k^2+49k^2
AC^2= 113k^2

AC =√113 k
According to the sine and cos function ratios,
it is written as
sinθ = AB/AC
= Opposite Side/Hypotenuse
= 8k/√113 k
= 8/√113
andcosθ=Adjacent Side/Hypotenuse
= BC/AC
= 7k/√113 k
= 7/√113
Now apply the values of sin function and cos function:

i)

(1 + sin θ)(1 - sin θ)/(1+cos θ)(1-cos θ)= 1-sin^2θ/1-cos^2θ
= cos^2θ/sin^2θ
=49/64

ii) cot^2 θ=cos^2θ/sin^2θ
=49/64


Thanks